How many numbers are there in the form of AABB which is a perfect square? Example: 7744 is in the form of AABB and it is square of 88.
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To begin with, let us note that the number we are looking at (namely A A B B ) is a number divisible by 1 1 , since A − A + B − B = 0 ≡ 0 ( m o d 1 1 ) . You may find further insight into the divisibility rules by 1 1 h e r e .
Let us say that A A B B = M 2 . Since A A B B ≡ 0 ( m o d 1 1 ) , note that A A B B can be factorized as A A B B = 1 1 n ⋅ R , where n ∈ N and R represents the other factors of A A B B .
Consequently, we see that M 2 must also be divisible by 1 1 (since M 2 = 1 1 n ⋅ R ). Let us also note that if a product of integers is divisible by a particular integer, then that must mean that at least one of the multipliers in the product must also be divisible with that integer. In this case, it is obvious that 1 1 ∣ M .
Therefore, we see that M is a multiple of 1 1 ! This actually does all the work for us, considering that there are only 7 multiples of 1 1 that form 4 -digit numbers and those are 3 3 , 4 4 , 5 5 , 6 6 , 7 7 , 8 8 and 9 9 .
At this point, we may easily fine the squares for each of them, which are 1 0 8 9 , 1 9 3 6 , 3 0 2 5 , 4 3 5 6 , 5 9 2 9 , 7 7 4 4 and 9 8 0 1 accordingly. From this, we can see that 7 7 4 4 is the only number satisfying this condition, which is ironically the square of 8 8 and also the example given to us by Tanmay.
Thus, the answer is 1 .